C2H6O + 3O2 —> 2CO2 + 3H2O
Answer:

Explanation:
When percentage composition is given, and asked for the empirical formula, it is simplest to assume 100 g of material. Thus,
Mass C = 40.92 g. Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C
Mass H = 4.58 g. Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H
Mass O = 54.50 g. Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O
Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.
Moles C = 3.41/3.41 = 1
Moles H = 4.58/3.41 = 1.34
Moles O = 3.41/3.41 = 1
Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3
Moles C = 1x3 = 3
Moles H = 1.34x3 = 4
Moles O = 1x3 = 3
Empirical Formula 
<span>Answer:
mol Al2O3 x 1 mol Al/ 2 mol Al2O3= .25 mol Al
The balanced equation tells us that it takes 4 moles of Al to produce 2 moles of Al2O3.
0.50 moles Al2O3 x (4 moles Al / 2 moles Al2O3) = 1.0 moles Al
1.0 moles Al x (27.0 g Al / 1 mole Al) = 27.0 g Al</span>
Answer:
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Explanation:
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It will slowly rotten and turn brown